Let $y=f(x)$ be an infinitely differentiable function on real numbers such that $f(0)$ is not equal to 0 and $d^n(y)∕dx^n$ not equal to zero at $x=0$ for $n=1,2,3,4$. If $$\lim_{x→0} {f(4x) + af(3x) + bf(2x) + cf(x) + df(0)\over x^4} $$ exists then find the value of $25a + 50b +100c + 500d$.

The answer is :


but I can't solve it.

  • $\begingroup$ We don't really need $f$ to be infinitely differentiable, just existence of fourth derivative at $x = 0$ is fine. But its OK if the question is giving more data than necessary. $\endgroup$ – Paramanand Singh May 13 '16 at 7:26
  • $\begingroup$ It would help if you can upvote my answer (which you have accepted) because although it is perfectly correct it is suffering from a downvote by some user. $\endgroup$ – Paramanand Singh May 15 '16 at 7:55
  • $\begingroup$ I am new here so I am short of reputation points. Once I cross them I would do it. $\endgroup$ – Harry Karwasra May 15 '16 at 9:53

Well the right tool to use here is Taylor's series and not L'Hospital's Rule. Note that we don't need anything more than the existence of $f^{(4)}(0)$ and the fact that all of $f(0), f'(0), f''(0), f'''(0), f^{(4)}(0)$ are non-zero.

We have via Taylor's series $$f(x) = f(0) + xf'(0) + \frac{x^{2}}{2}f''(0) + \frac{x^{3}}{6}f'''(0) + \frac{x^{4}}{24}f^{(4)}(0) + o(x^{4})\tag{1}$$ Replacing $x$ by $2x, 3x, 4x$ respectively we get \begin{align} f(2x) &= f(0) + 2xf'(0) + 2x^{2}f''(0) + \frac{4x^{3}}{3}f'''(0) + \frac{2x^{4}}{3}f^{(4)}(0) + o(x^{4})\tag{2}\\ f(3x) &= f(0) + 3xf'(0) + \frac{9x^{2}}{2}f''(0) + \frac{9x^{3}}{2}f'''(0) + \frac{27x^{4}}{8}f^{(4)}(0) + o(x^{4})\tag{3}\\ f(4x) &= f(0) + 4xf'(0) + 8x^{2}f''(0) + \frac{32x^{3}}{3}f'''(0) + \frac{32x^{4}}{3}f^{(4)}(0) + o(x^{4})\tag{4} \end{align} Hence we have the numerator of the limit expression given by \begin{align} Nr &= f(4x) + af(3x) + bf(2x) + cf(x) + df(0)\notag\\ &= f(0)(1 + a + b + c + d) \notag\\ &\,\,\,\,+ xf'(0)\left(4 + 3a + 2b + c\right)\notag\\ &\,\,\,\,+x^{2}f''(0)\left(8 + \frac{9a}{2} + 2b + \frac{c}{2}\right)\notag\\ &\,\,\,\,+x^{3}f'''(0)\left(\frac{32}{3} + \frac{9a}{2} + \frac{4b}{3} + \frac{c}{6}\right)\notag\\ &\,\,\,\,+x^{4}f^{(4)}(0)\left(\frac{32}{3} + \frac{27a}{8} + \frac{2b}{3} + \frac{c}{24}\right) + o(x^{4})\notag \end{align} From the above it is clear that the limit in question exists only when the following equations are satisfied simultaneously: \begin{align} a + b + c + d &= -1\tag{5a}\\ 3a + 2b + c &= -4\tag{5b}\\ 9a + 4b + c &= -16\tag{5c}\\ 27a + 8b + c &= -64\tag{5d} \end{align} Solving these equations we get $$a = -4, b = 6, c = -4, d = 1$$ and thus the value of $$25a + 50b +100c + 500d$$ is $300$.

Update: Note that the problem can also be solved using L'Hospital's Rule but it needs more careful analysis. I wished to highlight this point in comments to another answer of this question but it was difficult to give all the details in a comment.

Since the limit $$L = \lim_{x \to 0}\frac{f(4x) + af(3x) + bf(2x) + cf(x) + df(0)}{x^{4}}\tag{6}$$ exists it is clear that as $x \to 0$ the numerator of the above fraction tends to $0$. Hence $f(0)(1 + a + b + c + d) = 0$ which implies that $$a + b + c + d = -1\tag{7}$$ Next we know that the fraction in the limit $(6)$ is of the form $0/0$ and hence there is a chance that L'Hospital Rule will work. On applying LHR we get the fraction $$\frac{4f'(4x) + 3af'(3x) + 2bf'(2x) + cf'(x)}{4x^{3}}\tag{8}$$ The numerator tends to $f'(0)(4 + 3a + 2b + c)$ and if $(4 + 3a + 2b + c) \neq 0$ then the above fraction tends to infinity as $x \to 0$. This would imply that the original limit $(6)$ will also be infinite. Thus we must have $$3a + 2b + c = -4\tag{9}$$ This means that the fraction in $(8)$ is also of the form $0/0$ (but we don't know if its limit exists or not). Hence there is a chance that further application of LHR might work. Doing this we get another fraction $$\frac{16f''(4x) + 9af''(3x) + 4bf''(2x) + cf''(x)}{12x^{2}}\tag{10}$$ The numerator tends to $f''(0)(16 + 9a + 4b + c)$ and if this is non-zero then the ratio in $(10)$ would tend to infinity and hence by LHR the ratio in $(8)$ would also tend to infinity. This would imply that the original limit in $(6)$ is also infinite which is not the case. Hence we must have $$9a + 4b + c = -16\tag{11}$$ In similar manner we can show that $$27a + 8b + c = -64\tag{12}$$ From $(7), (9), (11), (12)$ we can find values of $a, b, c, d$ and get the same answer $300$.

  • $\begingroup$ Would the downvoter care to comment?? $\endgroup$ – Paramanand Singh May 14 '16 at 4:25

Since the limit exists, denote it with $L$.

$$L = \lim_{x→0} {f(4x) + af(3x) + bf(2x) + cf(x) + df(0)\over x^4}$$

If we attempt to take the limit, we naively may think

$$L = {f(0)(1 + a + b + c + d)\over 0^4}$$

If the numerator is nonzero, then the expression diverges. Yet, $L$ exists. Therefore we expect the indeterminate form $\frac00$. The numerator must equal zero. Specifically, since $f(0)\ne 0$,

$$1 + a + b + c + d=0$$

Since the limit gives an indeterminate form, we employ L'Hopital as well:

$$L = \lim_{x→0} {4f'(4x) + 3af'(3x) + 2bf'(2x) + cf'(x) \over 4x^3}$$

This time, with a bit of work we can conclude

$$4+3a+2b+c = 0$$

Can you take it from here?


If you're concerned about the application of L'Hopital, convince yourself that the above steps are justified recursively, as is the case with many applications of L'Hopital's rule, upon reaching the finite limit $L$. This post is clearly not a thorough proof, but a guide to the problem at hand.

  • 1
    $\begingroup$ The limit doesn't diverge, the expression does. $\endgroup$ – zhw. May 13 '16 at 3:51
  • $\begingroup$ Thanks. Will edit. That is more accurate phrasing. $\endgroup$ – zahbaz May 13 '16 at 3:53
  • $\begingroup$ this does not work as we don't know if the limit after applying L'Hospital's Rule exist or not. The proper mechanism is to use Taylor's series. There is a way to salvage this with more careful analysis of each limit obtained after applying L'Hospital's Rule, but a naive application of the rule as you have done is not correct. $\endgroup$ – Paramanand Singh May 13 '16 at 7:20
  • $\begingroup$ @ParamanandSingh Why wouldn't it? If $\lim_{x\rightarrow0}f(x)\ne0$, then the original limit must have reached an indeterminate form if $L$ is to exist. The conditions for L'Hopital can be inferred, and we can apply the rule successively until the denominator reaches 1. What alternative outcome could occur? $\endgroup$ – zahbaz May 13 '16 at 8:56
  • $\begingroup$ No! L'Hospital rule uses the existence of $\lim f'/g'$ to infer existence of $\lim f/g$. Here you know that $\lim f/g$ exists and you then also assume that $\lim f'/g'$ also exists. This is not the way L'Hospital works. $\endgroup$ – Paramanand Singh May 13 '16 at 9:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.